Integrals Involving Logarithmic Functions: Videos & Practice Problems

Learn to integrate rational functions using substitution. The integral of $\frac{1}{x}$ is $\ln \left(\right|x\left|\right)$ plus a constant. For complex functions like $\frac{4}{3+4x}$, set $u$ as the denominator, then integrate $\frac{1}{u}$. For trigonometric functions, use hints like multiplying by $\frac{\sec x+\tan x}{\sec x+\tan x}$ to simplify integration.

Integrals Resulting in Natural Logs

In calculus, understanding the relationship between derivatives and integrals is crucial. The derivative of the natural logarithm function, specifically ln(x), is given by the formula:

$$\frac{d}{dx} \ln(x) = \frac{1}{x}$$

This relationship allows us to derive a corresponding integral rule. By reversing the differentiation process, we can find the integral of 1/x. Thus, the integral can be expressed as:

$$\int \frac{1}{x} \, dx = \ln |x| + C$$

Here, C represents the constant of integration, and the absolute value is included to ensure the function is defined for all values of x, including negative values.

Additionally, 1/x can be rewritten as x^-1, which emphasizes that negative exponents indicate a reciprocal. This understanding is essential when applying the constant multiple rule in integration.

For example, when integrating 5/x, we can factor out the constant:

$$\int \frac{5}{x} \, dx = 5 \int \frac{1}{x} \, dx = 5 \ln |x| + C$$

Next, consider a more complex integral, such as:

$$\int \left(\frac{1}{x^2} + \frac{3}{x}\right) \, dx$$

This can be separated into two distinct integrals using the sum rule:

$$\int \frac{1}{x^2} \, dx + \int \frac{3}{x} \, dx$$

For the first integral, 1/x² can be rewritten as x^-2. Applying the power rule, we find:

$$\int x^{-2} \, dx = -x^{-1} = -\frac{1}{x}$$

For the second integral, we again apply the constant multiple rule:

$$\int \frac{3}{x} \, dx = 3 \int \frac{1}{x} \, dx = 3 \ln |x|$$

Combining these results, the final answer for the integral becomes:

$$-\frac{1}{x} + 3 \ln |x| + C$$

This process illustrates the application of integration rules, including the power rule and the constant multiple rule, while reinforcing the importance of recognizing the relationship between derivatives and integrals. Continued practice with various integrals will enhance understanding and proficiency in calculus.

Integrals Resulting in Natural Logs Example 1

To solve the indefinite integral of the expression \( \frac{x^3 e^x - 4x^2}{x^3} \, dx \), we can simplify the problem by breaking it down into two separate fractions. This allows us to handle each term more easily.

First, we rewrite the integral as follows:

\[\int \left( \frac{x^3 e^x}{x^3} - \frac{4x^2}{x^3} \right) \, dx\]

Upon simplifying, we notice that the \( x^3 \) in the numerator cancels with the \( x^3 \) in the denominator for the first term, and in the second term, \( x^2 \) cancels out two of the \( x^3 \) terms, leaving us with:

\[\int \left( e^x - \frac{4}{x} \right) \, dx\]

This integral is now much simpler to evaluate. The antiderivative of \( e^x \) is simply \( e^x \). For the second term, we apply the rule for integrating \( \frac{1}{x} \), which gives us \( -4 \ln |x| \). Therefore, we can combine these results:

\[\int \left( e^x - \frac{4}{x} \right) \, dx = e^x - 4 \ln |x| + C\]

Here, \( C \) represents the constant of integration. Thus, the final answer for the indefinite integral is:

\[e^x - 4 \ln |x| + C\]

Understanding how to break down complex integrals into simpler components is a valuable skill in calculus, allowing for easier evaluation of antiderivatives.

Find the indefinite integral.
${\int }_{}^{}\frac{3-y^2}{2y}dy$

• A. $\frac{3}{2}\ln \mid y\mid -\frac{1}{4}{y}^2+C$
• B. $\frac{3}{2}+\frac{1}{4y^2}+C$
• C. $\frac{3}{2}-\frac{1}{4y^2}+C$
• D. $\frac{3}{2}\ln \mid y\mid -\frac{1}{4y^2}+C$

Answer: (A)

Integrals Resulting in Natural Logs Example 2

To evaluate the definite integral from 1 to 5 of \(\frac{dT}{4T}\), we can start by rewriting the integral in a more familiar form. Recognizing that \(\frac{dT}{4T}\) can be expressed as \(\frac{1}{4} \cdot \frac{dT}{T}\), we can factor out the constant \(\frac{1}{4}\) from the integral:

\[\int_{1}^{5} \frac{dT}{4T} = \frac{1}{4} \int_{1}^{5} \frac{dT}{T}\]

Next, we apply the integral rule for \(\frac{1}{T}\), which states that:

\[\int \frac{dT}{T} = \ln |T| + C\]

Thus, the integral becomes:

\[\frac{1}{4} \left[ \ln |T| \right]_{1}^{5}\]

Since both bounds are positive, we can drop the absolute value signs. Now we evaluate the integral at the bounds:

\[\frac{1}{4} \left( \ln(5) - \ln(1) \right)\]

Knowing that \(\ln(1) = 0\), we simplify this to:

\[\frac{1}{4} \ln(5)\]

This expression represents the exact value of the definite integral. If a numerical approximation is required, using a calculator will yield approximately \(0.4\) for \(\frac{1}{4} \ln(5)\). This result provides a clear understanding of how to evaluate integrals involving logarithmic functions.

Evaluate the definite integral.
${\int }_2^e\frac{3+x}{x}dx$

• A. $4e-8\approx 2.87$
• B. $e-3\ln 2-1\approx -0.36$
• C. $e-3\approx -0.28$
• D. $1+e-3\ln 2\approx 1.64$

Answer: (D)

Find the area under the graph of $f\left(t\right)=\frac{t^2{e}^t+t}{t^2}$ from $t=1$ to $t=3$.

• A. $4e^3+\ln 3-2e\approx 76.00$
• B. $4e^3+\ln 3-2e-1\approx 75.00$
• C. $e^3-e+\ln 3\approx 18.47$
• D. $e^2+\ln 3\approx 8.49$

Answer: (C)

Integrals Involving Natural Logs: Substitution

To find the integral of a more complicated rational function, such as \(\frac{4}{3 + 4x}\), we can utilize substitution along with the integral of \(\frac{1}{x}\). The first step is to set \(u = 3 + 4x\). Consequently, the differential \(du\) becomes \(du = 4 \, dx\), or \(dx = \frac{du}{4}\). Substituting these into the integral transforms it into \(\int \frac{4}{u} \cdot \frac{du}{4} = \int \frac{1}{u} \, du\). The integral of \(\frac{1}{u}\) is \(\ln |u| + C\). Replacing \(u\) back with \(3 + 4x\), we arrive at the final result: \(\ln |3 + 4x| + C\).

For another example, consider the integral of \(\frac{2x + 4}{x^2 + 4x} \, dx\). Here, we again set \(u = x^2 + 4x\), leading to \(du = (2x + 4) \, dx\). This allows us to rewrite the integral as \(\int \frac{du}{u}\), which evaluates to \(\ln |u| + C\). Substituting back gives us \(\ln |x^2 + 4x| + C\).

In a more complex case, such as \(\frac{5}{x \ln(x^3)} \, dx\), we can choose a different substitution. Instead of using the entire denominator, we set \(u = \ln(x)\). Then, \(du = \frac{1}{x} \, dx\), allowing us to rewrite the integral as \(5 \int \frac{1}{u^3} \, du\). This can be expressed as \(5 \int u^{-3} \, du\), which evaluates to \(-\frac{5}{2u^2} + C\). Substituting back gives us the final answer: \(-\frac{5}{2(\ln(x))^2} + C\).

In summary, when dealing with integrals of rational functions, the key is to identify a suitable substitution that simplifies the integral into a recognizable form. If the initial choice of substitution does not yield a simpler integral, consider alternative substitutions or polynomial division to facilitate the integration process.

Find the indefinite integral.
${\int }_{}^{}\frac{1}{2x+5}dx$

• A. $\ln \mid 2x+5\mid +C$
• B. $\frac{1}{2}\ln \mid 2x+5\mid +C$
• C. $2x+5+C$
• D. $\frac{3}{2x^3+15x}+C$

Answer: (B)

Integrals Involving Natural Logs: Substitution Example 3

To find the indefinite integral of the function \(\frac{e^{2x}}{1 + e^{2x}} \, dx\), we can utilize a strategic substitution. Instead of letting \(u = 2x\), which complicates the derivative, we will set \(u\) equal to the entire denominator: \(u = 1 + e^{2x}\). This choice simplifies our calculations significantly.

Next, we compute the derivative \(du\). The derivative of \(u\) is given by:

\[du = 2e^{2x} \, dx\]

This means that we can express \(e^{2x} \, dx\) in terms of \(du\) as follows:

\[e^{2x} \, dx = \frac{1}{2} du\]

Now, substituting \(u\) and \(du\) into the integral, we have:

\[\int \frac{e^{2x}}{1 + e^{2x}} \, dx = \int \frac{e^{2x}}{u} \cdot \frac{1}{2} du = \frac{1}{2} \int \frac{1}{u} \, du\]

Applying the integral rule for \(\frac{1}{u}\), we find:

\[\frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln |u| + C\]

Substituting back for \(u\), we get:

\[\frac{1}{2} \ln |1 + e^{2x}| + C\]

Since \(1 + e^{2x}\) is always positive, we can drop the absolute value signs, leading to our final result:

\[\frac{1}{2} \ln(1 + e^{2x}) + C\]

This integral demonstrates the power of substitution in simplifying complex expressions, allowing us to arrive at a clear and concise solution.

Integrals Involving Natural Logs: Substitution Example 4

To find the indefinite integral of the function \(\frac{2}{x \ln x} \, dx\), we can utilize a substitution method. Initially, we consider the denominator \(x \ln x\) for substitution, but this approach complicates the integral due to the product rule involved in finding \(du\). Instead, we focus on the natural logarithm present in the integrand.

By setting \(u = \ln x\), we can derive \(du\) as follows: since the derivative of \(\ln x\) is \(\frac{1}{x}\), we have \(du = \frac{1}{x} \, dx\). This allows us to express \(dx\) in terms of \(du\): \(dx = x \, du\). Substituting \(u\) into the integral, we rewrite the original integral:

\[\int \frac{2}{x \ln x} \, dx = 2 \int \frac{1}{\ln x} \cdot \frac{1}{x} \, dx\]

Now, substituting \(u\) into the integral gives us:

\[2 \int \frac{1}{u} \, du\]

Applying the integral rule for \(\frac{1}{u}\), we find:

\[2 \ln |u| + C\]

Substituting back for \(u\), we have:

\[2 \ln |\ln x| + C\]

Since \(\ln x\) is always positive for \(x > 1\), we can drop the absolute value, leading to the final result:

\[2 \ln(\ln x) + C\]

This expression represents the indefinite integral of the given function, showcasing the relationship between logarithmic functions and their integrals.

Example 5: Integral of Tangent & Cotangent

To find the integral of tangent \( x \), we start by rewriting it in terms of sine and cosine. The integral can be expressed as:

\[\int \tan x \, dx = \int \frac{\sin x}{\cos x} \, dx\]

Next, we use substitution. Let \( u = \cos x \), which gives us \( du = -\sin x \, dx \). Rearranging this, we have \( -du = \sin x \, dx \). Substituting these into the integral, we get:

\[\int \tan x \, dx = -\int \frac{1}{u} \, du\]

This integral is straightforward to solve, yielding:

\[-\ln |u| + C\]

Substituting back for \( u \), we find:

\[-\ln |\cos x| + C\]

Using properties of logarithms, we can rewrite this as:

\[\ln |\sec x| + C\]

Thus, the integral of tangent \( x \) is:

\[\int \tan x \, dx = \ln |\sec x| + C\]

Now, let's consider the integral of cotangent \( x \). We rewrite cotangent in terms of sine and cosine:

\[\int \cot x \, dx = \int \frac{\cos x}{\sin x} \, dx\]

Using a similar substitution, let \( u = \sin x \), which gives \( du = \cos x \, dx \). This allows us to rewrite the integral as:

\[\int \cot x \, dx = \int \frac{1}{u} \, du\]

Integrating this, we obtain:

\[\ln |u| + C\]

Substituting back for \( u \), we have:

\[\ln |\sin x| + C\]

Therefore, the integral of cotangent \( x \) is:

\[\int \cot x \, dx = \ln |\sin x| + C\]

In summary, the integrals of the tangent and cotangent functions are:

\[\int \tan x \, dx = \ln |\sec x| + C\]

\[\int \cot x \, dx = \ln |\sin x| + C\]

Example 6: Integral of Secant & Cosecant

To find the integral of secant and cosecant functions, we can utilize a strategic approach involving substitution and manipulation of the integrands. Starting with the integral of secant, we express it as:

$$\int \sec x \, dx$$

To simplify this integral, we multiply by the fraction $$\frac{\sec x + \tan x}{\sec x + \tan x}$$, which is effectively multiplying by one. This gives us:

$$\int \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x} \, dx$$

Next, we set $$u = \sec x + \tan x$$. The derivative $$du$$ is calculated as:

$$du = (\sec x \tan x + \sec^2 x) \, dx$$

Noticing that the numerator matches $$du$$, we can rewrite the integral as:

$$\int \frac{1}{u} \, du$$

This integral evaluates to:

$$\ln |u| + C = \ln |\sec x + \tan x| + C$$

Thus, the integral of secant is:

$$\int \sec x \, dx = \ln |\sec x + \tan x| + C$$

Next, we turn our attention to the integral of cosecant:

$$\int \csc x \, dx$$

Similar to the previous case, we multiply by the fraction $$\frac{\csc x + \cot x}{\csc x + \cot x}$$, leading to:

$$\int \frac{\csc^2 x + \csc x \cot x}{\csc x + \cot x} \, dx$$

We set $$u = \csc x + \cot x$$, and find that:

$$du = -(\csc x \cot x + \csc^2 x) \, dx$$

To align with our integral, we multiply by -1, resulting in:

$$-\int \frac{1}{u} \, du$$

This evaluates to:

$$-\ln |u| + C = -\ln |\csc x + \cot x| + C$$

Thus, the integral of cosecant is:

$$\int \csc x \, dx = -\ln |\csc x + \cot x| + C$$

In summary, the integrals of secant and cosecant yield important results that can be used in various applications:

1. $$\int \sec x \, dx = \ln |\sec x + \tan x| + C$$

2. $$\int \csc x \, dx = -\ln |\csc x + \cot x| + C$$

Integrals Involving Natural Logs: Substitution Example 7

To find the indefinite integral of the function \(\frac{x^2 + 2x + 3}{x + 3} \, dx\), we start by recognizing that the numerator is a quadratic polynomial while the denominator is linear. When the degree of the numerator exceeds that of the denominator, polynomial long division is a useful technique to simplify the integrand.

We begin by dividing \(x^2 + 2x + 3\) by \(x + 3\). The first step in polynomial long division involves determining what to multiply \(x + 3\) by to obtain \(x^2\). We find that multiplying \(x + 3\) by \(x\) yields \(x^2 + 3x\). Subtracting this from the original polynomial results in:

\(x^2 + 2x + 3 - (x^2 + 3x) = -x + 3\).

Next, we bring down the remaining term, resulting in \(-x + 3\). We then determine what to multiply \(x + 3\) by to get \(-x\), which is \(-1\). Multiplying gives us \(-x - 3\). Subtracting this from \(-x + 3\) yields:

\(-x + 3 - (-x - 3) = 6\).

At this point, we have completed the division, resulting in:

\(x - 1 + \frac{6}{x + 3}\).

We can now rewrite the integral as:

\(\int \left( x - 1 + \frac{6}{x + 3} \right) \, dx\).

Next, we can integrate each term separately. The integral of \(x\) using the power rule is \(\frac{1}{2}x^2\). The integral of \(-1\) is \(-x\). For the term \(\frac{6}{x + 3}\), the integral results in \(6 \ln |x + 3|\). Therefore, combining these results, we have:

\(\int \left( x - 1 + \frac{6}{x + 3} \right) \, dx = \frac{1}{2}x^2 - x + 6 \ln |x + 3| + C\),

where \(C\) is the constant of integration. This expression represents the final answer for the indefinite integral.

Integrals Involving Natural Logs: Substitution Example 8

To compute the indefinite integral of the function x over (x - 2)^2, we can utilize a substitution method. We start by letting u = x - 2. Consequently, the differential du = dx, since the derivative of x - 2 is 1. This substitution simplifies our integral, but we still need to express x in terms of u. Rearranging gives us x = u + 2.

Substituting these values into the integral, we rewrite it as:

$$\int \frac{u + 2}{u^2} \, du$$

This expression can be split into two separate fractions:

$$\int \left( \frac{u}{u^2} + \frac{2}{u^2} \right) \, du = \int \left( \frac{1}{u} + 2u^{-2} \right) \, du$$

Now, we can integrate each term individually. The integral of 1/u is the natural logarithm:

$$\int \frac{1}{u} \, du = \ln |u|$$

For the second term, we apply the power rule:

$$\int 2u^{-2} \, du = -2u^{-1} = -\frac{2}{u}$$

Combining these results, we have:

$$\ln |u| - \frac{2}{u} + C$$

Finally, we substitute back for u to express our answer in terms of x:

$$\ln |x - 2| - \frac{2}{x - 2} + C$$

In summary, when integrating rational functions, choosing an appropriate substitution can simplify the process significantly. If one method does not yield results, it is beneficial to explore alternative approaches. Practice will enhance your ability to identify the most effective strategies for integration.

Evaluate the definite integral.
${\int }_1^6\frac{dt}{2+5t}$

• A. $\ln \left(\frac{32}{7}\right)\approx 1.52$
• B. $\ln \left(25\right)\approx 3.22$
• C. $\frac{1}{5}\ln \left(25\right)\approx 0.64$
• D. $\frac{1}{5}\ln \left(\frac{32}{7}\right)\approx 0.30$

Answer: (D)

Evaluate the definite integral.
${\int }_0^{\frac{\pi }{3}}\frac{sin\theta }{1+\cos \theta }d\theta$

• A. $\ln \left(\frac{4}{3}\right)\approx 0.29$
• B. $\ln \left(\frac{3}{4}\right)\approx -0.29$
• C. $\ln \left(2\right)-\cos \left(\frac{\pi }{3}\right)\ln (1+\cos \frac{\pi }{3})\approx 1.99\times {10}^{-4}$
• D. $\ln \left(2\right)-\cos \left(\frac{\pi }{3}\right)\ln (1+\cos \frac{\pi }{3})\approx 0.49$

Answer: (A)

Find the area under the graph of $f\left(x\right)=\frac{e^{-x}}{1+e^{-x}}$ from $x=0$ to $x=4$.

• A. $\ln \left(\frac{2}{1+e^{-4}}\right)\approx 0.67$
• B. $\ln \left(\frac{2e^{-4}}{1+e^{-4}}\right)\approx -3.33$
• C. $e\ln 2\approx 1.88$
• D. $\ln \left(\frac{1}{1+e^{-4}}\right)\approx -0.018$

Answer: (A)

The region between the curve $y=\frac{2}{\sqrt{x}}$ and the $x$-axis from $x=1$ to $x=3$ is revolved about the $x$-axis to form a solid. Find the volume of this solid. 

• A. $8\pi \ln \left(3\right)\approx 27.61$
• B. $4\pi \approx 12.56$
• C. $2\pi \ln \left(3\right)\approx 6.90$
• D. $4\pi \ln \left(3\right)\approx 13.81$

Answer: (D)

Solve the initial value problem given by $\frac{dy}{dx}=\frac{2}{x}+3,y\left(1\right)=4$.

• A. $y=2\ln \mid x\mid +3x+4$
• B. $y=2\ln \mid x\mid +3x$
• C. $y=2\ln \mid x\mid +3x+1$
• D. $y=3x+1$

Answer: (C)

Here’s what students ask on this topic:

What is the integral of 1/x and how is it derived?

The integral of $\frac{1}{x}$ is derived from the fact that the derivative of the natural logarithm function, $\ln \left(x\right)$, is $\frac{1}{x}$. By reversing this process, we find that the integral of $\frac{1}{x}$ is $\ln \left(\right|x\left|\right)$ plus a constant of integration, $c$. The absolute value is used to ensure the function is defined for all values of $x$ except zero. This rule is fundamental in calculus for integrating rational functions.

How do you integrate more complex rational functions using substitution?

To integrate complex rational functions, substitution is a powerful technique. For a function like $\frac{4}{(}$, set $u$ equal to the denominator, $3+4x$. Then, find $du$ by differentiating $u$, which gives $4dx$. Substitute these into the integral, transforming it into $\frac{1}{u}$ du, which integrates to $\ln \left(\right|u\left|\right)$ plus $c$. Finally, substitute back $u$ to express the integral in terms of $x$.

What is the integral of secant and how is it evaluated?

The integral of $\sec \left(x\right)$ is evaluated using a clever substitution. Multiply the integrand by $\frac{\sec }{\left(}$ over itself, which simplifies the integral to $\frac{1}{u}$ du where $u$ is $\sec \left(x\right)+\tan \left(x\right)$. This integrates to $\ln \left(\right|u\left|\right)$ plus $c$. Substituting back gives the result: $\ln \left(\right|\sec \left(x\right)+\tan \left(x\right)\left|\right)$ plus $c$.

How do you integrate the function 5/x ln(x)^3?

To integrate $\frac{5}{x}$, use substitution. Set $u$ to $\ln \left(x\right)$, making $du$ equal to $\frac{1}{x}$ dx. The integral becomes $5\frac{1}{u}$ du. This simplifies to $5(-\frac{1}{2}u^-2)$ plus $c$. Substitute back to get $-\frac{5}{2}\ln \left(x\right)^-2$ plus $c$.

What is the integral of cosecant and how is it evaluated?

The integral of $\csc \left(x\right)$ is evaluated similarly to secant. Multiply by $\frac{\csc }{\left(}$ over itself, simplifying the integral to $-\frac{1}{u}$ du where $u$ is $\csc \left(x\right)+\cot \left(x\right)$. This integrates to $-\ln \left(\right|u\left|\right)$ plus $c$. Substituting back gives the result: $-\ln \left(\right|\csc \left(x\right)+\cot \left(x\right)\left|\right)$ plus $c$.